 Explicit midpoint linear multistep method

For an ordinary differential equation $y^{\prime }=\Phi (t,y)$, this linear multistep method uses a centered derivative across the two-step interval $[t_{i-1},t_{i+1}]$, evaluated at its midpoint $t_i$. It requires two starting values, $y_0$ and $y_1$. For a sufficiently smooth exact solution, Taylor's theorem givesThe unscaled local residual is therefore $O(h^3)$ and the method has order two. Its first characteristic polynomial is $z^2-1$, with the simple roots $1,-1$, so it satisfies the root condition for a multistep method; the second root is a parasitic amplification root. This two-step method differs from the one-step midpoint method and the implicit midpoint rule.

For the special equation $G^{\prime }=g(t)$ and $G(a)=0$, an even number $n=2r$ of steps gives $y_n=2h{\sum }_{j=1}^{r}g(t_{2j-1})$, independently of the odd-index starting value. It is the composite midpoint quadrature rule on $r$ intervals of width $2h$. An odd final index depends on $y_1$, which must be specified separately.